Let P be a point on the hyperbola H: x^2 9- y^2 4=1, in the first quadrant such that the area of triangle formed by P and the two foci of H is 2√13.…

Plotting the diagram of given data we get,

Given equation of hyperbola is $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{4}=1$9x2​−4y2​=1.

$$\Rightarrow {a}^{2}=9,{b}^{2}=4$$⇒a2=9,b2=4

We know that, ${b}^{2}={a}^{2}({e}^{2}-1)$b2=a2(e2−1)

$$\Rightarrow {e}^{2}=1+\frac{{b}^{2}}{{a}^{2}}$$⇒e2=1+a2b2​

$$\Rightarrow {e}^{2}=1+\frac{4}{9}$$⇒e2=1+94​

$$\Rightarrow {e}^{2}=\frac{13}{9}$$⇒e2=913​

$$\Rightarrow e=\frac{\sqrt{13}}{3}$$⇒e=313​​

Now, ${S}_{1}{S}_{2}=2ae$S1​S2​=2ae

$$\Rightarrow {S}_{1}{S}_{2}=2\times 3\times \sqrt{\frac{13}{3}}$$⇒S1​S2​=2×3×313​​

$$\Rightarrow {S}_{1}{S}_{2}=2\sqrt{13}$$⇒S1​S2​=213​

It is given that, area of $\Delta P{S}_{1}{S}_{2}$ΔPS1​S2​ is $2\sqrt{13}$213​.

$$\Rightarrow \frac{1}{2}\times \beta \times {S}_{1}{S}_{2}=2\sqrt{13}$$⇒21​×β×S1​S2​=213​

$$\Rightarrow \frac{1}{2}\times \beta \times 2\sqrt{13}=2\sqrt{13}$$⇒21​×β×213​=213​

$$\Rightarrow \beta =2$$⇒β=2

Since, $(\alpha ,\beta )$(α,β) lies on the hyperbola,

$$\Rightarrow \frac{{\alpha }^{2}}{9}-\frac{{\beta }^{2}}{4}=1$$⇒9α2​−4β2​=1

$$\Rightarrow \frac{{\alpha }^{2}}{9}-1=1$$⇒9α2​−1=1

$$\Rightarrow {\alpha }^{2}=18$$⇒α2=18

$$\Rightarrow \alpha =3\sqrt{2}$$⇒α=32​

Distance of $P$P from origin is given by,

$$OP=\sqrt{{\alpha }^{2}+{\beta }^{2}}$$OP=α2+β2​

$$\Rightarrow OP=\sqrt{18+4}$$⇒OP=18+4​

$$\Rightarrow OP=\sqrt{22}$$⇒OP=22​

$$\Rightarrow O{P}^{2}=22$$⇒OP2=22